X^y=y^x

The problem I looked into was the following: What are all the solutions to the equation x y = y x ? As it turns out, it's kind of wierd to look at the equation in that form. So I reduced it to a more functional-friendlyproblem: x y = yx y Log[x] = x Log[y] x Log@xD

=

y LogAyE

Which, interpreted graphically, asks: At which points does the graph of f(x) =

x Log@xD

not pass the horizontal line test? Or, in other words, what values of

x Log@xD

can be obtained with two different x values? First, lets take a look at the plot of f(x) =

x Log@xD

.

f@x_D=xLog@xD x PlotBf@xD,8x,0,8<,PlotRange®8-2,2ã<,PlotLabel®"fHxL=

"F Log@xD

x Log@xD

x fHxL= Log@xD 5 4 3 2 1

2

4

6

8

-1 -2

We can see that as when x < 1, the y values are all negative, and since all those for x > 1 are positive there are clearly no solutions for x < 1. So, the fun stuff happens when x > 1, as we can see from the graph. We can do some functional analysis on f(x) to see where the minimum occurs:

f'(x) =

Log@xD-1 Log@xD2

f'(x) = 0 when x = ã, and f(ã) = ã. (A pretty cool property of this function, but not the point.) Also note that Limit@ f HxL, x ® ¥D = ¥ and Limit@ f HxL, x ® 1D = ¥. Thus, for x between 1 and ã and for x > ã, f(x) obtains all real values greater than ã, and since the function is continuous it hits every value greater than ã exactly twice. (Exactly twice because the derivative is zero only once, and thus increasing on (ã,¥).)

2

x^y=y^x.nb

So, how will we find these values? I'll spare you the details, but below is a data chart of the solutions and some analyses of the data. The data points are (x,y) as in the equation above- they satisfy the relation

x Log@xD

=

y LogAyE

.

data = {{2.7182818109319746`, 2.718281828459045`}, {1.494853040773493`, 7.484072499348272`}, 11.540109238744765`}, {1.2427541304507346`, 15.772583628556744`}, {1.1946027884970856`, 20.189544570480795`}, 24.774145588106872`}, {1.1396478582309841`, 29.50861329037053`}, {1.1224309841455844`, 34.37746138841561`}, 39.3676546161954`}, {1.0982537247599835`, 44.46824757622764`}, {1.0894347175396006`, 49.66998953325085`}, 54.964991232882966`}, {1.07583273032812`, 60.346461322837314`}, {1.070476542122498`, 65.80850114761421`}, 71.34594494393073`}, {1.0617562025232532`, 76.95423457041677`}, {1.0581591911744188`, 82.62932042004714`}, 88.36758227377531`}, {1.0520923201889816`, 94.16576546201654`}, {1.0495104584156205`, 100.02092888621424`}, 105.9304023180605`}, {1.0471726645909594`, 105.9304023180605`}, {1.0450458650029162`, 111.89175102569726`}, 117.90274623982417`}, {1.0413203759367855`, 123.96134031540868`}, {1.0396796844869993`, 130.0656457003125`}, 136.21391701450037`}, {1.0367606237467366`, 142.40453568955348`}, {1.0354565168463647`, 148.63599673009975`}, 154.9068972452273`}, {1.0331076143205038`, 161.21592646529717`}, {1.0320461756675223`, 167.56185701245283`}}

fit = Fit@data, 8xLog@xD<, xD Show@ListPlot@dataD, Plot@fit, 8x, 1, 1.5 "Data & BestFit"D Plot@f@fitD - f@xD, 8x, 1.00001, 1.5<, PlotRange -> 8-ã,ã<, PlotLabel -> f@fitD - f@xDD

4.713900059471978` x Log@xD

Data & BestFit 150

100

50

1.4

1.6

1.8

2.0

{1.3239138060132911`, {1.1625596258027329`, {1.1090115878602411`, {1.0820724439573632`, {1.0658283223646892`, {1.054958626149123`, {1.0471726645909594`, {1.0431027077651465`, {1.0381643849000988`, {1.0342418112062484`,

x^y=y^x.nb

4.7139 x logHxL logJ

4.7139 x N logHxL

-

3

x logHxL

2

1

1.1

1.2

1.3

1.4

1.5

-1

-2

The first graph here is a plot of the data and a best fit line of the data. The best fit line is of the form

cx Log@xD

, because in my experimentationthat turned out to be

the closest estimation to the data. I was disappointed to find that there are no elementary functions that approximate the data, and here's why:

We're essentially looking for a function y(x) so that f(y(x)) = f(x), but y(x) isn't simply x. If we could find such a function, (it would be defined for all x > 1), then it would catagorize all the pairs of points (x, f(x)) and (n, y(n)) where f(x) = f(n), which is the stated goal. I began by algebraically making a guess for y(x) and then seeing if

x Log@xD

was actually algebraically equivalent to

yHxL LogAyHxLE

. For example, I tried the function y(x) =

x Log@xD

, or y(x) = f(x), and it came scarily close

to working. x

x Log@xD

=

Log@xD x

LogB

Log@xD

F

Simplifying rapidly, x Log@xD

=ã

ã x = xã Which of course isn't always true, but certainly looks pretty good.

The conclusion is that the data obtained here can't be classified by any elementary algebraic functions. Going back to the original equation, x y = y x , we can actually observe some cool stuff that can now be explained:

4

x^y=y^x.nb

ManipulateAPlotA8x^y,y^x<,8x,1,6<, PlotRange->81,Hã+.3L^Hã+.3L+1<,PlotLabel® "xy vs. yx as y varies"E,8y,1,ã+1<E

y

x y vs. y x as y varies

25

20

15

10

2

3

4

5

6

Since I didn't mention it before, there is one trivial solution where x = y. You can see that one in the graph above, as well as the other non-trivial solution that I worked on. We notice that when x = ã, the two graphs are tangent- they touch, but don't pass through. The reason they always have 2 intersections(instead of 0) is that the polynomial (blue), which starts out lower than the exponential, is growing faster than the exponential (red) at first. But, asymptotically, the exponential grows much faster than the polynomial- so it they have another intersection point. I'd maximize this window to get the best view of their intersections.